The 'Moreover' part follows immediately from commuting \(\frac{\Theta _{ba}^{j}}{j!}, ~\frac{\Theta _{ac}^{k_a}}{k_a!},\)  in the RHS of (REF ), using the commutations relations (REF ) and the fact that \(\Theta _{ba}\)  restricted to the copy of \(S^{(k_a,k_b)}\hookrightarrow M^{(k_a,k_b)}\)  is zero.
Regarding the main part, first note that both sides of (REF ) map \(M^{(k_a,k_b)}\)  to \(M^{(k_a,j,k_b-j)}.\)  The hook length formula [1]} guarantees that \(S^{(k_a,k_b)}\)  appears in the latter \(S_{k_a+k_b}\) -module with multiplicity \(1.\)  Since \(\Theta _{xy}\)  are morphisms of \(S_{k_a+k_b}\) -modules by Lemma REF (REF ), it must happen the the expressions on the LHS and RHS of (REF ) differ by a multiplicative scalar, that may also be zero.
Since these two expressions are maps of modules, in order to calculate this scalar it is enough to apply both sides of (REF ) to one element of \(S^{(k_a,k_b)}\)  and compare the results.
In order to describe a non zero element \(e_T\)  we shall use the well known construction of Specht module. Consider the Young diagram \(D_{k_a,k_b}=(k_a,k_b).\)  Recall that a Young tableau \(T\)  corresponds to a basis element \(e_T\)  of \(M^{(k_a,k_b)},\)  in which the entries of the first row correspond to the locations of \({\text{\bfseries \ttfamily a}}\)  while those of the second row to the locations of \({\text{\bfseries \ttfamily b}}.\)  Consider the Standard Young tableau \(T\)  in which the first row contains the elements \(1,2,\ldots , k_b,2k_b+1,2k_b+2,\ldots , k_a.\)  Define
\(v_T=\sum _{\pi }(-1)^{sgn(\pi )}e_{\pi \cdot T},\) 
where the summation is taken over \(\pi \in S_{k_a+k_b}\)  which preserve the columns of \(T.\)  \(v_T\)  can be written more explicitly as
\(\sum _{R \in R_{T}}(-1)^{s(R)}e_R,\) 
where \(R_{T}\)  is the set of Young tableau for \(D_{k_a,k_b}\)  for which the rightmost elements in the first row are \(2k_b+1,\ldots ,k_a,\)  in the \(j-\) th column for \(m\le k_b\)  the entries are \(m,k_b+m\)  in some order, and \(s(R)=\)
\(|\lbrace k_b<m\le 2k_b|~m~\text{appears in the first row of } R\rbrace |\\=|\lbrace m\le k_b|~m~\text{appears in the second row of }R\rbrace |.\) 
Let \(D_{k_a,j,k_b-j}\)  we the Young diagram with rows of length \(k_a,j,k_b-j.\)  The corresponding Young tableaux index the standard basis of \(M^{(k_a,j,k_b-j)},\)  where the first row indicates the locations of \({\text{\bfseries \ttfamily c}}\) , the row of length \(j\)  the locations of \({\text{\bfseries \ttfamily a}}\)  and the remaining row the locations of \({\text{\bfseries \ttfamily b}}.\) 
We first calculate
\(\frac{\Theta _{ba}^{j}}{j!}\frac{\Theta _{ac}^{k_a}}{k_a!} v_T=\sum _{Q\in Q_T}(-1)^{s_r(Q)}e_Q,\) 
where \(Q\)  is a Young tableau of \(D_{k_a,j,k_b-j},~e_Q\)  the corresponding basis element, \(Q_T\)  is the set of Young tableaux for which the indices greater than \(2k_b\)  are in the first row, and for each \(m\le k_b\)  exactly one index in \(\lbrace m,k_b+m\rbrace \)  appears in the first row.
The sign
\(s_r(Q)=|\lbrace m\le k_b|~m~\text{does not appear in the first row of }Q\rbrace |.\) 
The proof is straightforward. The application of \(\Theta _{ac}^{k_a}/k_a!\)  changes all \({\text{\bfseries \ttfamily a}}\)  to \({\text{\bfseries \ttfamily c}}\)  and does nothing else. Then \(\Theta _{ba}^j/j!\)  changes exactly \(j\)  of the \({\text{\bfseries \ttfamily b}}\) 's to \({\text{\bfseries \ttfamily a}}.\)  Note that any \(Q\in Q_T\)  comes from a single \(R\in R_T,\)  obtained by uniting the second and third row of \(Q\)  into one row. From here also the translation of \(s(-)\)  to \(s_r(-)\)  is immediate.
We now calculate the application of the operator on the LHS of (REF ) on \(v_T.\) 
We first show
\(v^{\prime }_T:=\frac{\Theta _{ab}^{i}}{i!} v_T=\sum _{X\in X_T}(-1)^{s(X)}e_X,\) 
where \(X_T\)  is the collection of Young tableaux for the Young diagram \(D_{k_a-i,k_b+i}\)  for which in the \(m-\) th column for \(m\le k_b\)  the entries are \(m,k_b+m\)  in some order, and \(s(X)\)  equals again
\(|\lbrace k_b<m\le 2k_b|~m~\text{appears in the first row of } X\rbrace |\\=|\lbrace m\le k_b|~m~\text{appears in the second row of }X\rbrace |.\) 
For the proof, define, for a given Young tableau \(X\)  of shape \((k_a-i,k_b+i)\)  the set \(P(X)\)  of Young tableaux \(R\in R_T\)  such that
\(X\)  appears with non zero coefficient in the representation of \(\frac{\Theta _{ab}^i}{i!} e_R\)  according to the standard basis.
It is easy to see that the coefficient, if non zero, is \(1.\) 
Thus, the coefficient of \(e_X\)  in \(\frac{\Theta _{ab}^i}{i!} v_T\)  is
\(\sum _{R\in P(X)}(-1)^{s(R)}.\) 
When \(X\in X_T\)  it is straight forward to see that \(P(X)\)  is the singleton \(R\)  defined by removing the boxes containing the highest \(i\)  values in the row of \({\text{\bfseries \ttfamily b}}\)  and adding them to the row of \({\text{\bfseries \ttfamily a}}\) . This argument also explains the sign \(s(X)=s(T).\)
For any \(X\notin X_T,\)  with non empty \(P(X),\)  there is at least one \(m\le k_b\)  such that both \(m\)  and \(m+k_b\)  appear in the row of \({\text{\bfseries \ttfamily b}}.\) 
Let \(m_\star \)  be the least such \(m.\)  Define an involution \(\iota :P(X)\rightarrow P(X)\)  as follows. For \(R\in P(X)\)  exactly one of \(m_{\star },k_b+m_{\star }\)  appears in the row of \({\text{\bfseries \ttfamily b}}.\)  Then \(\iota (R)\)  is the same tableau, except that if in \(R~m_{\star }\)  is in the row of \({\text{\bfseries \ttfamily b}},\)  in \(\iota (R)\)  it will be \(k_b+m_{\star },\)  and vice versa.
\(\iota \)  clearly maps \(P(X)\)  to itself, and is an involution. Moreover, by the definition of the sign \(s(R),\)  it is easily seen to be sign reversing,
\(s(\iota (R))=1-s(R).\) 
Thus,
\(\sum _{R\in P(X)}(-1)^{s(R)}=&\sum _{R\in \iota (P(X))}(-1)^{s(R)}=\sum _{R\in P(X)}(-1)^{s(\iota (R))}\\=&\sum _{R\in P(X)}(-1)^{1-s(R)}=-\sum _{R\in P(X)}(-1)^{s(R)}.\) 

Therefore, the coefficient of \(e_X,\)  which is sum of (REF ), vanishes, proving (REF ) holds, since
The next step is to calculate
\(\frac{\Theta _{bc}^{i+j}}{(i+j)!}\frac{\Theta _{ac}^{k_a-i-j}}{(k_a-i-j)!} v^{\prime }_T=\binom{k_a-k_b}{i}\sum _{Q\in Q_T}(-1)^{s_l(Q)}e_Q,\) 
where \(Q_T\)  is as above, but
\(s_l(Q)=|\lbrace m\le k_b|~m~\text{appears in the row of ${\text{\bfseries \ttfamily b}}$ in $Q$ or $k_b+m$ in the row of ${\text{\bfseries \ttfamily a}}$ in }Q\rbrace |.\) 
This proves the lemma, since \(s_l(Q)s_r(Q)=(-1)^{j}.\)  Indeed, modulo \(2,\)  the sum of
\(|\lbrace m\le k_b|~m~\text{appears in the row of ${\text{\bfseries \ttfamily b}}$ in $Q$} \text{ or $k_b+m$ appears in the row of ${\text{\bfseries \ttfamily a}}$ in }Q\rbrace |\) 
and
\(|\lbrace m\le k_b|~m~\text{does not appear in the first row of }Q\rbrace |\) 
equals
\(|\lbrace m\le k_b|~\text{either $m$ or $k_b+m$ appears in the row of ${\text{\bfseries \ttfamily a}}$}\rbrace |.\) 
This can be seen by defining three sets of \(m\le k_b\) , so that in the first \(m\)  and \(m+k_b\)  are in the rows of \({\text{\bfseries \ttfamily b}}\)  and \({\text{\bfseries \ttfamily c}}\)  respectively, in the second they're in the rows of \({\text{\bfseries \ttfamily c}}\)  and \({\text{\bfseries \ttfamily a}}\)  respectively, and in the third they're in the rows of \({\text{\bfseries \ttfamily a}}\)  and \({\text{\bfseries \ttfamily c}}\)  respectively, and noticing that each of the the above three sets whose cardinality we use is a disjoint union of another pair of the new sets.
Since for any \(Q\in Q_T\)  the row of \({\text{\bfseries \ttfamily a}}\)  is made of \(j\)  elements, all at most \(2k_b,\)  and for any \(m\le k_b\)  at most one of \(\lbrace m,k_b+m\rbrace \)  belongs to the row of \({\text{\bfseries \ttfamily a}}\)  (because by our definition of \(Q_T\) , exactly one index in \(\lbrace m,k_b + m\rbrace \)  appears in the first row, that of \({\text{\bfseries \ttfamily c}}\) , so either only the other one appears in the row of \({\text{\bfseries \ttfamily a}}\) , or neither of them appear in this row), the last cardinality is exactly the length of the row of \({\text{\bfseries \ttfamily a}},\)  that is \(j.\)
We are left with proving (REF ). First note that \(Q_T\)  is made of tableaux such that all elements greater than \(2k_b\)  are in the row of \({\text{\bfseries \ttfamily c}},\)  while for any \(m\le k_b,\)  exactly one of \(\left\lbrace m,m+k_b\right\rbrace \)  is in that row. In addition, exactly \(j\)  elements from \(\lbrace 1,2,\ldots , 2k_b\rbrace \)  appear in the row of \({\text{\bfseries \ttfamily a}}.\) 
The proof is similar to the proof of (REF ), and uses an involution argument again. For \(Q\)  of shape \((k_a,j,k_b-j)\)  let \(P^{\prime }(Q)\)  be the collection of elements \(X\in X_T\)  such that \(e_Q\)  appears in the representation of \(\frac{\Theta _{bc}^{i+j}}{(i+j)!}\frac{\Theta _{ac}^{k_a-i-j}}{(k_a-i-j)!}e_X\)  in the standard basis.
The coefficient of \(e_Q\)  in \(\frac{\Theta _{bc}^{i+j}}{(i+j)!}\frac{\Theta _{ac}^{k_a-i-j}}{(k_a-i-j)!} v^{\prime }_T\) 
is \(\sum _{X\in P^{\prime }(Q)}(-1)^{s(X)}\) .
Suppose \(Q\notin Q_T.\)  Then from the pigeon hole principle there must be an \(m\le k_b\)  such that both \(m,\)  and \(k_b+m\)  appear in the first row. Let \(m_{\star }\)  be the minimal such \(m\)  (for the given \(Q\) ).
Define an involution \(\iota ^{\prime }:P^{\prime }(Q)\rightarrow P^{\prime }(Q)\)  as follows. For any \(X\in P^{\prime }(Q)\)  either the \(m_\star \)  appears in the row of \({\text{\bfseries \ttfamily a}}\)  and \(k_b+m_\star \)  in that of \({\text{\bfseries \ttfamily b}},\)  or the opposite. In both cases \(\iota ^{\prime }(X)\)  is defined by moving \(m_{\star },k_b+m_{\star }\)  to the row in which they did not appear.
Again \(s(X)=1-s(\iota ^{\prime }(X)),\)  and again we see that the coefficient of \(e_X\)  in the LHS of (REF ) vanishes.
For \(Q\in Q_T,\)  on the other hand, the set \(P^{\prime }(Q)\)  is easily described: For any \(m\le k_b\)  which appears in the first row of \(Q\) , if \(m+k_b\)  is in the row of \({\text{\bfseries \ttfamily a}},\)  then for any \(X\in P^{\prime }(Q)\)  it holds that \(m\)  appears in the row of \({\text{\bfseries \ttfamily b}},\)  and \(m+k_b\)  appears in the row of \({\text{\bfseries \ttfamily a}}.\) 
Similarly,

if \(Q\)  has \(m\le k_b\)  in the first row and \(m+k_b\)  in the row of \({\text{\bfseries \ttfamily b}},\)  then \(X\in P^{\prime }(Q)\)  has \(m\)  in the row of \({\text{\bfseries \ttfamily a}}\)  and \(m+k_b\)  in the row of \({\text{\bfseries \ttfamily b}}\) ,

if \(Q\)  has \(k_b<m+k_b\le 2k_b\)  in the first row and \(m\)  in the row of \({\text{\bfseries \ttfamily a}},\)  then \(X\in P^{\prime }(Q)\)  has \(m\)  in the row of \({\text{\bfseries \ttfamily a}}\)  and \(m+k_b\)  in the row of \({\text{\bfseries \ttfamily b}}\) ,

if \(Q\)  has \(k_b<m+k_b\le 2k_b\)  in the first row and \(m\)  in the row of \({\text{\bfseries \ttfamily b}},\)  then \(X\in P^{\prime }(Q)\)  has \(m\)  in the row of \({\text{\bfseries \ttfamily b}}\)  and \(m+k_b\)  in the row of \({\text{\bfseries \ttfamily a}}\) .

These are the only four possibilities for \(Q\in Q_T\) .
Since \(s(X)\)  is defined using only the locations of the first \(2k_b\)  elements, it is the same for all \(X \in P^{\prime }(Q)\) , and by the last argument it is equal to \(s_l(Q)\) .
The first row in \(Q\)  contains \(k_a-k_b\)  entries that are greater than \(2k_b\) , and these could come from either the \({\text{\bfseries \ttfamily a}}\)  row or the \({\text{\bfseries \ttfamily b}}\)  row in \(X\in P^{\prime }(Q)\) , without any restriction. Therefore, we can get each element in \(P^{\prime }(Q)\)  uniquely by assigning the entries up to \(2k_b\)  as determined, and then choosing some \(i\)  elements greater than \(2k_b\)  from the first row of Q to put in the \({\text{\bfseries \ttfamily b}}\)  row of \(X\) .
Thus, \(|P^{\prime }(Q)|=\binom{k_a-k_b}{i},\)  since they all have the same sign \(s_l,~\) (REF ) follows.
